Question

A small candle,2.5cm in size is placed at 27cm in front of a concave mirror of radius of curvature 36cm.At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image.If the is moved closer to the mirror, how would the screen have to be moved?

Answer 1

Size of the candle,h=2.5cm
Image size=h'
Object distance,u=-27cm
The radius of curvature of the concave mirror, R=-36cm
The focal length of the concave mirror,ƒ=$\frac{R}{2}$=-18cm
Image distance=v
The image distance can be obtained using the mirror formula:$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$=$\frac{1}{-18}-\frac{1}{-27}$=$-3+\frac{2}{54}$=$-\frac{1}{54}$
Therefore, the screen should be placed 54cm away from the mirror to obtain a sharp image. The magnification of the image is given as m=$\frac{h'}{h}=-\frac{v}{u}$
∴h'=$-\frac{v}{u}$×h=-($-\frac{54}{-27}$)×2.5=-5cm
The height of the candle's image is 5cm. The negative sign indicates that the image is inverted and virtual. If the candle is moved closer to the mirror, then the screen will have to move away from the mirror in order to obtain the image.